Friction

MOTION OF A BODY ON ROUGH HORIZONTAL SURFACE

Applying a Horizontal Force :
Consider a body of mass "m" lying on a horizontal surface of coefficient of static friction "\(\mu_s\)". Let a horizontal force "F" is applied on the body as shown in the figure is increased gradually, If 'F' is too small, the body does not slide.
\(\Rightarrow\)Frictional force is static friction and it is equal to force that tends to move the body. (F_tm)
but F_tm= F\(\Rightarrow\)f = F to bring it into motion. For the body to come in to motion the applied force "F" must be at least equal to the limiting friction "f_L".
                          

            F=\(\mu_s\)N; F=\(\mu_s\)mg     \( \left( {\because N = mg} \right) \)
Consider a body of mass "m" placed on a rough horizontal surface of coefficient of kinetic friction "\(\mu_k\)". A horizontal force "F" greater than limiting friction is applied on it so that the body comes in to motion. Now kinetic friction acts on it in a direction opposite to its motion. 
If "N" is the normal reaction, then the kinetic friction is given by 
            f_k=\(\mu_k\)N=\(\mu_k\) mg
If "a" is the acceleration produced in the body, he resultant force acting on the body is given by
            \( \begin{gathered} f_R = F - f_k \Rightarrow ma = F - f_k \hfill \\ \therefore a = \frac{{F - f_k }} {m} = \frac{{F - \mu _k mg}} {m} \hfill \\ \end{gathered} \)
             

Note : The distance travelled and velocity acquired in a given interval of time  "t" can be obtained from the
kinematic equations \( S = ut + \frac{1} {2}at^2 \) and v = u+at.
Note : A force "F" just enough to set the block into motion is applied on the block. This by definition is equal to maximum static friction (\( f_L = \mu _s mg \)). If this force is continued, even after the block starts moving, the body now has to overcome kinetic friction (\( f_k = \mu _k mg \)) which is less than the applied force. Thus, there is a net resultant force which produces acceleration in the body. Then,
                                     

         \( F = f_L = \mu _s mg \)
         \( f_k = \mu _k mg \) 
         \( F_R = f_L - f_k \) 
         \( \begin{gathered} F_R = \left( {\mu _s mg - \mu _k mg} \right) \hfill \\ ma = \left( {\mu _s - \mu _k } \right)mg \hfill \\ \therefore a = \left( {\mu _s - \mu _k } \right)g \hfill \\ \end{gathered} \)